Physics - Rotation Question with Solution | TestHub

The total moment of inertia of the system about the axis $X{X}^{\prime }$ is the sum of the moment of inertia of the rod and the moment of inertia of the two point masses.

Moment of inertia of the uniform rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is:
$I_{\text{rod}}=\frac{M{L}^2}{12}$

The two point masses, each of mass $m$, are located at the ends of the rod, which is at a distance of $L/2$ from the axis of rotation.
Moment of inertia of the two point masses is:
$I_{\text{masses}}=m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2=2m\left(\frac{L^2}{4}\right)=\frac{m{L}^2}{2}$

Total moment of inertia $I$ is:
$I=I_{\text{rod}}+I_{\text{masses}}$

$I=\frac{M{L}^2}{12}+\frac{m{L}^2}{2}$

$I=\frac{M{L}^2+6m{L}^2}{12}$

$I=\frac{(M+6m)L^2}{12}$

Answer: $\frac{(M+6m)L^2}{12}$